HDU 5303 Delicious Apples

Delicious Apples

Time Limit: 5000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Problem Description

There are n apple trees planted along a cyclic road, which is L metres long. Your storehouse is built at position 0 on that cyclic road.
The ith tree is planted at position xi, clockwise from position 0. There are ai delicious apple(s) on the ith tree.

You only have a basket which can contain at most K apple(s). You are to start from your storehouse, pick all the apples and carry them back to your storehouse using your basket. What is your minimum distance travelled?

1≤n,k≤105,ai≥1,a1+a2+...+an≤105
1≤L≤109
0≤x[i]≤L

There are less than 20 huge testcases, and less than 500 small testcases.

 

Input

First line: t, the number of testcases.
Then t testcases follow. In each testcase:
First line contains three integers, L,n,K.
Next n lines, each line contains xi,ai.

 

Output

Output total distance in a line for each testcase.

 

Sample Input

210 3 22 28 25 110 4 12 28 25 10 10000

 

Sample Output

1826

 刚看这个题的时候头都大了，感觉贪心起来好复杂啊。情况太多，不是很好处理。但是多列几种情况就会发现，绕圈的话最多只需要绕一次，而且还必须是左半圈的苹果数加上右半圈的苹果数小于K时才有可能需要绕圈（自己举两个例子就明白了）。所以对所有需要绕圈情况枚举判断看看是不是更优即可。因为苹果总数不超过1e5，所以可以将苹果离散化，这样更好处理。用dpl[]和dpr[]分别表示取到某个苹果所需要的距离，这里需要用到贪心，想一想再不考虑其他情况下怎么取苹果才最优。例如两个苹果树1  k+1和2 k， 明显是要优先将远处的苹果取为一篮子，依据这个规则更新dp数组即可。代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <iostream>#include <algorithm>#include <functional>#define maxn 100010#define ll long longusing namespace std;long long dpl[maxn],dpr[maxn],dis[maxn];vector<int> vl,vr;int main(){        ios::sync_with_stdio(false);        int t;        cin>>t;        while(t--)        {                int l,n,k,cnt=1;                vl.clear();                vr.clear();                cin>>l>>n>>k;                for(int i=0;i<n;i++)                {                        int a,b;                        cin>>a>>b;                        for(int i=0;i<b;i++)                        {                                dis[cnt++]=a;                                if(a<=(l/2))                                {                                        vl.push_back(a);                                }                                else                                {                                        vr.push_back(l-a);                                }                        }                }                sort(vl.begin(),vl.end());                sort(vr.begin(),vr.end());                dpl[0]=0;                dpr[0]=0;                for(int i=0;i<vl.size();i++)                {                        if(i+1<=k)                        {                                dpl[i+1]=vl[i];                        }                        else                        {                                dpl[i+1]=vl[i]+dpl[i+1-k];                        }                }                for(int i=0;i<vr.size();i++)                {                        if(i+1<=k)                        {                                dpr[i+1]=vr[i];                        }                        else                        {                                dpr[i+1]=vr[i]+dpr[i+1-k];                        }                }                long long ans=dpl[vl.size()]*2+dpr[vr.size()]*2;                for(int i=0;i<=k&&i<=vl.size();i++)                {                        if(dpl[vl.size()-i]*2+dpr[max(0,(int)vr.size()-(k-i))]*2+l<ans)                        {                                ans=dpl[vl.size()-i]*2+dpr[max(0,(int)vr.size()-(k-i))]*2+l;                        }                }                cout<<ans<<endl;        }        return 0;}